What Happens to the Puck's Speed and Trajectory after the Weight Drops?

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When the lower weight drops after the string is burned, the puck's speed and trajectory radius change due to the conservation of energy and momentum. Initially, the centripetal force is provided by the tension in the string, equal to the weight of the 200g mass. After the weight drops, the centripetal force is now equal to the tension from the remaining 100g weight. The initial kinetic energy of the puck must equal the final kinetic energy plus the potential energy change as the system adjusts. The discussion highlights the importance of energy conservation in solving the problem, even when momentum is the focus.
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A 100g puck is revolving in a 20-cm-radius circle on a frictionless table. The string passes through a hole in the center of the table and is tied to two 200g weights.

Suppose a flame burns through the string and causes the lower weight to fall off while the puck is revolving. What will be the puck's speed and the radius of its trajectory after the weight drops?


I was able to find the speed prior to the flame burning through the string. But after that, I have no clue how to solve for the radius and speed.
 
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It is a trick question.
 
Why is it a trick question?

I would have thought what would happen is the tangential velocity would be maintained while radius would vary such that centripetal acceleration would equal to the gravitational force acting on the 100g weight left?
 
Hi 2rip,

Suppose the initial radius is r1 (given). Afterward, let the radius be r2.

Initially, centripetal force = tension in string = weight of 200g.

Afterward, centripetal force== tension in string =weight of 100g

Initial KE of the puck = Final KE + increase of PE of 100g wt to get lifted by (r2-r1).

From these 3 eqns, you can solve it.
 
Ah, so you mean that the two 100 gram weights are hanging in tandem below the table and that the string between the two is burned off?

I agree with Shooting star that energy should be conserved since no external force acts on the system when the string is burned.
 
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Ohh thanks! That makes so it so much clearer. I didn't use energy to solve ><
 
How about this same question but in a chapter concerning momentum only, energy has not been covered yet.
 
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